Limits, Continuity and Differentiability

The value of $\displaystyle\lim_{x\to0}\frac1{x^3}\int_0^x\frac{t\ln(1+t)}{t^4+4}\,dt$ is

Let f be a real-valued function defined on the interval $(0,\infty)$ by $f(x)=\ln x+\displaystyle\int_0^x\sqrt{1+\sin t}\,dt$. Then which of the following statement(s) is (are) true?

Let
$$L=\lim_{x\to0}\frac{a-\sqrt{a^2-x^2}-\dfrac{x^2}4}{x^4},\qquad a>0.$$
If $L$ is finite, then

For the function
$$f(x)=x\cos\frac1x,\qquad x\ge1,$$

Match the statements/expressions given in Column I with the values given in Column II.

Let $p(x)$ be a polynomial of degree 4 having extremum at $x=1,2$ and
$$\lim_{x\to0}\left(1+\frac{p(x)}{x^2}\right)=2.$$
Then the value of $p(2)$ is

Let
$$g(x)=\frac{(x-1)^n}{\log\cos^m(x-1)};\quad0<x<2,\ m\text{ and }n\text{ are integers},\ m\ne0,\ n>0,$$
and let $p$ be the left hand derivative of $|x-1|$ at $x=1$. If $\displaystyle\lim_{x\to1^+}g(x)=p$, then

Let $f$ and $g$ be real valued functions defined on interval $(-1,1)$ such that $g''(x)$ is continuous, $g(0)\ne0$, $g'(0)=0$, $g''(0)\ne0$, and $f(x)=g(x)\sin x$.

STATEMENT-1: $\displaystyle\lim_{x\to0}\left[g(x)\cot x-g(0)\operatorname{cosec}x\right]=f''(0)$.

and

STATEMENT-2: $f'(0)=g(0)$.

Let $f(x)$ be differentiable on the interval $(0,\infty)$ such that $f(1)=1$, and
$$\lim_{t\to x}\frac{t^2f(x)-x^2f(t)}{t-x}=1$$
for each $x>0$. Then $f(x)$ is

$$\lim_{x\to\frac\pi4}\frac{\displaystyle\int_2^{\sec^2x}f(t)\,dt}{x^2-\frac{\pi^2}{16}}$$
equals

In the following $[x]$ denotes the greatest integer less than or equal to $x$.

Match the functions in Column I with the properties in Column II and indicate your answer by darkening the appropriate bubbles in the 4 × 4 matrix given in the ORS.